Assignment 28

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12:12:56 4.1.42 (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.

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RESPONSE --> a = 1, b = -2, c = -3 Since a = 1>0, the graph opens up. The x-coordinate of the vertex is x = -b/2a = -(-2)/2(1) = 1 The y-coordinate of the vertex is f( -b/2a) = f(1) = 1^2 - 2(1) - 3 = 0 Thus the vertex is (1,0) The axis of symmetry is x = 1 The discriminant is b^2 - 4ac = (-2)^2 - 4(1)(-3) = 16 x-intercept is: x^2 -2x -3 = 16 (x-3)(x+1) = 16 x= 19 ; x= 15 y-intercept is: f(16) (16)^2 - 2(16) - 3 = 221 Domain : All real numbers Range: {y|y >= 0} f is increasing on (15, infinity) f is decreasing on ( infinity, 15)

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12:13:35 The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3. The graph of this quadratic function will open upwards, since a > 0. The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4). The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the quadratic formula. Here will will factor to get (x - 3) ( x + 1) = 0 so that x - 3 = 0 OR x + 1 = 0, giving us x = 3 OR x = -1. So the x intercepts are (-1, 0) and (3, 0). The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3).

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RESPONSE --> the point needed to be (1,4) not (1,0)

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12:23:50 4.1.67 (was 4.1.50). What are your quadratic functions whose x intercepts are -5 and 3, for the values a=1; a=2; a=-2; a=5?

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RESPONSE --> f(x) = 1(x - (-5))(x-3) = 1(x+5)(x-3) = 1(x^2 +2x - 15)= x^2 +2x - 15 f(x) = 2 (x+5)(x-3) = 2(x^2 + 2x - 15) = 4x^2 + 4x - 30 f(x) = -2 (x^2 + 2x -15) = -2x^2 -4x + 30 f(x) = 5(x^2 + 2x - 15) = 5x^2 + 10x - 150

5 * 15 = 75, not 150; otherwise fine.

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12:24:00 Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3). If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15. If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30. If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30. If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.

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RESPONSE --> I think I got it correct.

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12:25:14 Does the value of a affect the location of the vertex?

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RESPONSE --> yes, because if you change the last number it will change too.

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12:25:22 In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1.

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RESPONSE --> ok

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12:25:27 The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following: For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12. For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24. For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24. For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60. So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60).

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RESPONSE -->

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12:28:54 4.1.78 (was 4.1.60). A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?

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RESPONSE --> 1500 x 250

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12:29:55 05-08-2006 12:29:55 ** If the farmer uses 2000 meters of fence, and fences x feet along the highway, then he have 2000 - x meters for the other two sides. So the dimensions of the rectangle are x meters by (2000 - x) / 2 meters. The area is therefore x * (2000 - x) / 2 = -x^2 / 2 + 1000 x. The graph of this function forms a downward-opening parabola with vertex at x = -b / (2 a) = -1000 / (2 * -1/2) = 1000. At x = 1000 the area is -x^2 / 2 + 1000 x = -1000^2 / 2 + 1000 * 1000 = 500,000, meaning 500,000 square meters. Since this is the 'highest' point of the area vs. dimension parabola, this is the maximum possible area. **

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NOTES -------> What I was explaining was that no matter what you come up with it wil be equal on two sides and equal on the smaller of the two sides.

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12:29:57 ** If the farmer uses 2000 meters of fence, and fences x feet along the highway, then he have 2000 - x meters for the other two sides. So the dimensions of the rectangle are x meters by (2000 - x) / 2 meters. The area is therefore x * (2000 - x) / 2 = -x^2 / 2 + 1000 x. The graph of this function forms a downward-opening parabola with vertex at x = -b / (2 a) = -1000 / (2 * -1/2) = 1000. At x = 1000 the area is -x^2 / 2 + 1000 x = -1000^2 / 2 + 1000 * 1000 = 500,000, meaning 500,000 square meters. Since this is the 'highest' point of the area vs. dimension parabola, this is the maximum possible area. **

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RESPONSE -->

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12:33:35 4.1.101 (was 4.1.80). A rectangle has one vertex on the line y=10-x, x>0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area that can be enclosed by the rectangle.

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RESPONSE --> A(x) = -x^2 + 10x A = -b/2a = -10.2(-1) = 10/2 = 5 A(5) = -(5)^2 + 10(5) = -25 + 50 = 25 The largest area is 25

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12:33:42 ** The dimensions of the rectangle are x and y = 10 - x. So the area is area = x ( 10 - x) = -x^2 + 10 x. The vertex of this rectangle is at x = -b / (2 a) = -10 / (2 * -1) = 5. Since the parabola opens downward this value of x results in a maximum area, which is -x^2 + 10 x = -5^2 + 10 * 5 = 25. **

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RESPONSE --> correct

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Again, overall you appear to be doing OK here. A little more detail in self critiques might be beneficial; let me know if you have questions.